
How do you apply the evaluation theorem to evaluate the integral $3tdt$ over the interval $\left[ 0,3 \right]$ ?
Answer
540k+ views
Hint: Integration is the summation of all the smaller parts. There are types of integration. One is definite integration and the other one is indefinite integration. Definite Integration is the summation of a function over a definite interval in which the function is continuous. It has a fixed and defined interval. While indefinite integration does not have a fixed interval. We are just supposed to integrate the given function to find its anti – derivative.
Complete step by step answer:
The Evaluation theorem states the following:
If $f$ is a continuous function on $\left[ a,b \right]$ then $\int\limits_{a}^{b}{f\left( x \right)dx}=F\left( a \right)-F\left( b \right)$ where $F$ is the anti – derivative of $f$.
$\dfrac{d}{dx}F\left( x \right)=f\left( x \right)$.
So now we have to integrate $3tdt$ in the interval $\left[ 0,3 \right]$ . From this we can conclude that this is definite integration.
$\Rightarrow \int\limits_{0}^{3}{3tdt}$
We know the integral of ${{x}^{n}}$. It is $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$
Upon integrating it, we get the following:
$\Rightarrow \left[ \dfrac{3{{t}^{2}}}{2} \right]_{0}^{3}$
Now according to the theorem , let us substitute the upper limit and the lower limit. Upon doing so, we get the following:
$\begin{align}
& \Rightarrow \dfrac{3}{2}\left[ {{\left( 3 \right)}^{2}}-{{\left( 0 \right)}^{2}} \right] \\
& \Rightarrow \dfrac{3}{2}\times 9 \\
& \Rightarrow \dfrac{27}{2} \\
\end{align}$
$\therefore $After applying the evaluation theorem, the value of the integral $3tdt$ over the interval $\left[ 0,3 \right]$ is $\dfrac{27}{2}$.
Note:It is very important to know all the theorems and properties of both definite integration and indefinite integration. We should know the integrals of all the basic functions as knowing these would allow us to quickly solve the question. There are also some standard general ways to do integration which needs to be practiced. There are some general substitutions which need to be remembered.
Complete step by step answer:
The Evaluation theorem states the following:
If $f$ is a continuous function on $\left[ a,b \right]$ then $\int\limits_{a}^{b}{f\left( x \right)dx}=F\left( a \right)-F\left( b \right)$ where $F$ is the anti – derivative of $f$.
$\dfrac{d}{dx}F\left( x \right)=f\left( x \right)$.
So now we have to integrate $3tdt$ in the interval $\left[ 0,3 \right]$ . From this we can conclude that this is definite integration.
$\Rightarrow \int\limits_{0}^{3}{3tdt}$
We know the integral of ${{x}^{n}}$. It is $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$
Upon integrating it, we get the following:
$\Rightarrow \left[ \dfrac{3{{t}^{2}}}{2} \right]_{0}^{3}$
Now according to the theorem , let us substitute the upper limit and the lower limit. Upon doing so, we get the following:
$\begin{align}
& \Rightarrow \dfrac{3}{2}\left[ {{\left( 3 \right)}^{2}}-{{\left( 0 \right)}^{2}} \right] \\
& \Rightarrow \dfrac{3}{2}\times 9 \\
& \Rightarrow \dfrac{27}{2} \\
\end{align}$
$\therefore $After applying the evaluation theorem, the value of the integral $3tdt$ over the interval $\left[ 0,3 \right]$ is $\dfrac{27}{2}$.
Note:It is very important to know all the theorems and properties of both definite integration and indefinite integration. We should know the integrals of all the basic functions as knowing these would allow us to quickly solve the question. There are also some standard general ways to do integration which needs to be practiced. There are some general substitutions which need to be remembered.
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